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olga55 [171]
2 years ago
13

Which of the following clusters of orbitals could form a shape similar to that shown here (Figure 3) in the valence shell of an

isolated atom or one about to enter into bonding with other atoms?
A) five sp³d
B) three sp² and one p orbital

Chemistry
1 answer:
pav-90 [236]2 years ago
5 0
"three sp²" is the one cluster of orbitals among the following choices given in the question that <span>could form a shape similar to that shown here (Figure 3) in the valence shell of an isolated atom or one about to enter into bonding with other atoms. The correct option among all the options given in the question is the second option.</span>
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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive
lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as <u>the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction depends on the temperature, so it must always be specified.

<u>We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene.</u> The constant Kc for the above reaction is,

Kc = \frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}

According to the law of ideal gases,  

PV = nRT  

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,  

[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}

[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}

So, the molar concentration of ethene (C₂H₄) will be,

[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}

Then, according to the law of ideal gases,

P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L}  x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0
1 year ago
Explain on the chemical structural basis why the products of the saponification reaction are soluble in water while the starting
aev [14]

Answer:

The description is outlined in subsection downwards and according to the query given.

Explanation:

  • Saponification seems to be a procedure that requires the conversion or transformation of fat, grease, or lipid by either the intervention of heating a mixture of aqueous alkali towards soap as well as an alcoholic. Soaps contain fatty acid salts, however, mono-fatty acids contain carbon atoms, such as sodium palmitate. Therefore, throughout the water, individuals were indeed soluble.
  • However, on another hand, owing to large hydrocarbon strings, triglycerides do not partake in hydrogen bonding. Therefore in water, they aren't dissolved.
7 0
1 year ago
Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher
cluponka [151]

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber  increases to 1693 cm¹.

6 0
2 years ago
A flask contains 1/3 mole of h2 and 2/3 mol of he. Compare the force on the wall per impact of h2 relative to that for he.
densk [106]

The force on the wall is actually the pressure exerted by gas molecules

Higher the pressure more the force exerted on the walls of container

The pressure depends upon the number of molecules of a gas

In a mixture of gas the pressure depends upon the mole fraction of the gas

As given the mole fraction of He is more than that of H2 therefore He will exert more pressure on the wall

The ratio of impact will be

H2 / He = 2/3 / 1/3 = 2: 1

8 0
1 year ago
Read 2 more answers
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
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